Trig Integration Discrepancy in Fourier Series Analysis

[NewtonianAlch]

Homework Statement

This question is part of Fourier Series in Circuit Analysis. There were fairly straightforward integrals which I calculated and confirmed using MAPLE to be correct, however the book gives somewhat different answers. I would presume that what I did was correct and the solutions manual made an error, however since it's a fairly large question with answers being carried forward I want to make doubly sure. Sorry about the size of the images, I will remove them after the problem

This is the integral essentially, the definite integral from 2 to 4 is left out because it's zero,

f(t) = 5 for 0 < t < 1
f(t) = 10 for 1 < t < 2

http://img189.imageshack.us/img189/1379/dsc0007ob.jpg

Homework Equations

cos (Pi/2) = (-1)$^{\frac{n-1}{2}}$

cos (Pi) = (-1)$^{n}$

The Attempt at a Solution

My answer came to this:

http://img542.imageshack.us/img542/7613/dsc0008ryz.jpg

EDIT: cos(nPi/2) goes to (-1)^n/2 - still doesn't reconcile my answers with the book though.

The MAPLE output was:

$5\,{\frac {1+\cos \left( 1/2\,n\pi \right) -2\,\cos \left( n\pi \right) }{n\pi }}$

The answer in the book was (last line before the table):

http://img600.imageshack.us/img600/9624/dsc0009cm.jpg

As you can imagine, because the answers are different, the values in the table are going to be different and hence whatever I have to plot afterwards will be different.

 

[gneill]

$(-1)^{integer}$ cannot produce any zeros, it can only produce +1 or -1. So it doesn't replace $cos(n \pi / 2)$.

 

[NewtonianAlch]

I should have been clearer, cos (n*Pi/2) is replaced by (-1)^n/2

So if n = 3, I'm guessing that term is ignored because you can't compute that. At least that's the identity they gave in the book.